Question

What is the Cell Potential of the electrochemical cell in Which the cell reaction is: Pb2+ + Cd → Pb + Cd2+ ; Given that Eocell = 0.277 volts, temperature = 298K, [Cd2+] = 0.02M, and [Pb2+] = 0.2M.Immersive Reader

Answer 1

Answer:

Nernst equation

Explanation:

Ecell = Eocell-0.059/n *log[anode/cathode]

E cell = .277 - 0.059/2 * log [.02/.2]           ( anode = cd 2+)  (cathode = pb2+)

on solving E cell = 0.3065 V

Answer 2

Given:

The electrochemical cell in Which the cell reaction is: Pb2+ + Cd → Pb + Cd2+; Given that Eocell = 0.277 volts, temperature = 298K, [Cd2+] = 0.02M, and [Pb2+] = 0.2M.

To Find:

The Cell Potential of the electrochemical cell in Which the cell reaction is: Pb2+ + Cd → Pb + Cd2+.

Solution:

To find the cell potential we will follow the following steps:

As we know,

The nearest equation for the cell at 298K is

$Ecell = Eocell- \frac{0.0591}{n} log \frac{product \: concentration}{reactant \: concentration}$

Here, n is the number of electrons exchanged which is equal to 2.

$Ecell = Eocell- \frac{0.0591}{2} log \frac{product \: concentration}{reactant \: concentration}$

$Ecell = 0.277- \frac{0.0591}{2} log \frac{0.02}{0.2}$

$Ecell = 0.277- \frac{0.0591}{2} \frac{1}{10}$

As we know log 1/10 = -1.

Now,

$Ecell = 0.277 + \frac{0.0591}{2}$

$Ecell = 0.277 + 0.02955$

$Ecell = 0.30volt$

Henceforth, the Cell Potential of the electrochemical cell in Which the cell reaction is: Pb2+ + Cd → Pb + Cd2+ is 0.30V.